A spring of negligible mass has force constant k = 1700 N/m . Part A How far mus

Question

A spring of negligible mass has force constant k = 1700 N/m .

Part A

How far must the spring be compressed for an amount 3.20 J of potential energy to be stored in it?

Express your answer using two significant figures.

Part B

You place the spring vertically with one end on the floor. You then drop a book of mass 1.40 kg onto it from a height of 0.500 m above the top of the spring. Find the maximum distance the spring will be compressed.

Take the free fall acceleration to be 9.80 m/s2 . Express your answer using two significant figures.

Explanation / Answer

part A:

potential energy of the spring=0.5*spring constant*compression^2

if compression is x m,

then 0.5*1700*x^2=3.2

==>x=0.061357 m=6.1357 cm

in two significant figure , spring has to be compressed by 6.14 cm

part B:

initial potential energy of the spring=0(as spring in relaxed state)

initial potential energy of the book=mass*g*height=1.4*9.8*0.5=6.86 J

hence initial total mechanical energy=6.86 J

when the spring is compressed by maximum distance,speed of the book=0==>kinetic energy of the book=0

assume that maximum compression is d m

then potential energy of the book=-1.4*9.8*d=-13.72*d

potential energy of the spring=0.5*1700*d^2

then total energy=850*d^2-13.72*d

using conservation of energy principle and equating total initial and final energy:

850*d^2-13.72*d=6.86

==>850*d^2-13.72*d-6.86=0

solving for d, we get d=0.0982 m=9.82 cm

hence the spring will be compressed maximum by 9.82 cm

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